化简cos(π/4-a)cos(π/4+a)

结果为 1/2cos2a
cos（α+β）=cosαcosβ-sinαsinβ
cos（α-β）=cosαcosβ+sinαsinβ

解:
cos(π/4-a)cos(π/4+a)=cos(a-π/4)cos(a+π/4)
=cos(a-(π/2-π/4))cos(a+π/4)
=cos((a+π/4)-π/2)cos(a+π/4)
=sin(a+π/4)cos(a+π/4)
=1/2sin(2a+π/2)
=1/2cos2a

该式：cos(a+b)*cos(a-b)=cos²a-sin²b
该题：
原式=cos²π/4-sin²a
=1/2-sin²a
=(1/2)(1-2sin²a)
=(1/2)cos2a
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cos(π/4-a)cos(π/4+a)
=(cos(π/4)*cosa + sin(π/4)*sina) * (cos(π/4)*cosa - sin(π/4)*sina
=√2/2 * (cosa+ sina) * √2/2 * (cosa- sina)
=1/2 * (cos²a - sin²a)
=(cos2a)/2

解法1：套用公式：cosα·cosβ=(1/2)[cos(α+β)+cos(α-β)]有：cos(π/4-a)cos(π/4+a)=(1/2)[cos(π/4-a+π/4+a)cos(π/4-a-π/4-a)]=(1/2)[cos(π/2)+cos(-2a)]=(1/2)[0+cos(2a)]=(1/2)cos(2a)解法2：cos(π/4-...

cos(π/4-a)
=sin[π/2-(π/4-a)]
=sin(π/4+a)
cos(π/4-a)cos(π/4+a)
=sin(π/4+a)cos(π/4+a)
=(1/2)sin(π/2+2a)
=(1/2)cos2a